How many distinct triangles can be formed for which m∠E = 64°, g = 9, and e = 10? triangle(s)

Mathematics

1 answer:

Shalnov [3]2 years ago

5 0

Using the law of sines, it is found that 1 triangle can be formed with the given conditions.

<h3>What is the law of sines?</h3>

Suppose we have a triangle in which:

The length of the side opposite to angle A is a.

The length of the side opposite to angle B is b.

The length of the side opposite to angle C is c.

The lengths and the sine of the angles are related as follows:

$\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}$

Hence, for this problem, we have that the relation is:

$\frac{\sin{64^\circ}}{10} = \frac{\sin{G}}{9}$

$\sin{G} = 0.9\sin{64^\circ}$

$\sin{G} = 0.8089$

$G = \arcsin{0.8089}$

G = 54º

A triangle can have one angle of 64º and another of 54º, hence 1 triangle can be formed with the given conditions.

More can be learned about the law of sines at brainly.com/question/25535771

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(a) $f'(x) = \frac{d}{dx}[\frac{lnx}{x}]$
Using the quotient rule:
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(b) <em>Deduce:
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<em>
Soln:</em> Since x = e is the greatest value, then f(e) ≥ f(x) > f(0)
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Now, since x > 0, then we don't have to worry about flipping the signs when multiplying by x.
$\frac{x}{e} \geq lnx$
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Taking the exponential to both sides will cancel with the natural logarithmic function in the right hand side to produce:
$e^{x} \geq e^{ln(x^{e}})$
$e^{x} \geq x^{e}$ , as required.